在 STRING_AGG 中生成 DISTINCT 值

Question

我在 SQL Server 2017 中使用 STRING_AGG 函数。我想创建与 COUNT(DISTINCT <column>) 相同的效果。我试过 STRING_AGG(DISTINCT <column>,',') 但这不是合法的语法。

我想知道是否有 T-SQL 解决方法。这是我的样本：

WITH Sitings 
  AS
  (
    SELECT * FROM (VALUES 
      (1, 'Florida', 'Orlando', 'bird'),
      (2, 'Florida', 'Orlando', 'dog'),
      (3, 'Arizona', 'Phoenix', 'bird'),
      (4, 'Arizona', 'Phoenix', 'dog'),
      (5, 'Arizona', 'Phoenix', 'bird'),
      (6, 'Arizona', 'Phoenix', 'bird'),
      (7, 'Arizona', 'Phoenix', 'bird'),
      (8, 'Arizona', 'Flagstaff', 'dog')
    ) F (ID, State, City, Siting)
  ) 
SELECT State, City, COUNT(DISTINCT Siting) [# Of Types], STRING_AGG(Siting,',') Animals
FROM Sitings 
GROUP BY State, City

上面的结果如下：

+---------+-----------+--------------+-------------------------+
|  State  |   City    | # Of Types   |         Animals         |
+---------+-----------+--------------+-------------------------+
| Arizona | Flagstaff |            1 | dog                     |
| Florida | Orlando   |            2 | dog,bird                |
| Arizona | Phoenix   |            2 | bird,bird,bird,dog,bird |
+---------+-----------+--------------+-------------------------+

输出正是我想要的，只是我希望为 Phoenix Arizona 列出的串联 "Animals" 是不同的，如下所示：

+---------+-----------+--------------+--------------------+
|  State  |   City    | # Of Types   |      Animals       |
+---------+-----------+--------------+--------------------+
| Arizona | Flagstaff |            1 | dog                |
| Florida | Orlando   |            2 | dog,bird           |
| Arizona | Phoenix   |            2 | bird,dog           |
+---------+-----------+--------------+--------------------+

有什么想法吗？

当我使用更大的真实数据集时，我收到关于 "Animals" 列超过 8000 个字符的错误。

我认为我的问题与this one相同，只是我的示例简单得多。

Answer 1

这是一种方法。

由于您还需要非重复计数，因此只需将行分组两次即可完成。第一个 GROUP BY 将删除重复项，第二个 GROUP BY 将产生最终结果。

WITH
Sitings
AS
(
    SELECT * FROM (VALUES 
    (1, 'Florida', 'Orlando', 'bird'),
    (2, 'Florida', 'Orlando', 'dog'),
    (3, 'Arizona', 'Phoenix', 'bird'),
    (4, 'Arizona', 'Phoenix', 'dog'),
    (5, 'Arizona', 'Phoenix', 'bird'),
    (6, 'Arizona', 'Phoenix', 'bird'),
    (7, 'Arizona', 'Phoenix', 'bird'),
    (8, 'Arizona', 'Flagstaff', 'dog')
    ) F (ID, State, City, Siting)
)
,CTE_Animals
AS
(
    SELECT
        State, City, Siting
    FROM Sitings
    GROUP BY State, City, Siting
)
SELECT
    State, City, COUNT(1) AS [# Of Sitings], STRING_AGG(Siting,',') AS Animals
FROM CTE_Animals
GROUP BY State, City
ORDER BY
    State
    ,City
;

结果

+---------+-----------+--------------+----------+
|  State  |   City    | # Of Sitings | Animals  |
+---------+-----------+--------------+----------+
| Arizona | Flagstaff |            1 | dog      |
| Arizona | Phoenix   |            2 | bird,dog |
| Florida | Orlando   |            2 | bird,dog |
+---------+-----------+--------------+----------+

如果您仍然收到有关超过 8000 个字符的错误消息，请在 STRING_AGG.

之前将值转换为 varchar(max)

类似

STRING_AGG(CAST(Siting AS varchar(max)),',') AS Animals

Answer 2

只需使用 sub-query

WITH Sitings 
      AS
      (
        SELECT * FROM (VALUES 
          (1, 'Florida', 'Orlando', 'bird'),
          (2, 'Florida', 'Orlando', 'dog'),
          (3, 'Arizona', 'Phoenix', 'bird'),
          (4, 'Arizona', 'Phoenix', 'dog'),
          (5, 'Arizona', 'Phoenix', 'bird'),
          (6, 'Arizona', 'Phoenix', 'bird'),
          (7, 'Arizona', 'Phoenix', 'bird'),
          (8, 'Arizona', 'Flagstaff', 'dog')
        ) F (ID, State, City, Siting)
      ) 

    select State,City,count(*) as [# Of Types],STRING_AGG(Siting,',') AS Animals from 
    (
      SELECT State, City, Siting
    FROM Sitings 
    GROUP BY State, City,Siting
    ) as T  group by State,City

http://sqlfiddle.com/#!18/ba4b8/11

  State     City    # Of Types  Animals
Arizona     Flagstaff   1   dog
Florida     Orlando     2   bird,dog
Arizona     Phoenix     2   bird,dog

Answer 3

当然是很晚才回复。

这是另一种方式。 STRING_AGG(Siting,',') 中的 Siting 可以被子查询以返回 SITTING 的 DISTINCT 列表，其中分组键匹配 STATE 和 CITIES。

WITH Sitings 
      AS
      (
        SELECT * FROM (VALUES 
          (1, 'Florida', 'Orlando', 'bird'),
          (2, 'Florida', 'Orlando', 'dog'),
          (3, 'Arizona', 'Phoenix', 'bird'),
          (4, 'Arizona', 'Phoenix', 'dog'),
          (5, 'Arizona', 'Phoenix', 'bird'),
          (6, 'Arizona', 'Phoenix', 'bird'),
          (7, 'Arizona', 'Phoenix', 'bird'),
          (8, 'Arizona', 'Flagstaff', 'dog')
        ) F (ID, State, City, Siting)
      ) 

SELECT 
    S.State, 
    S.City, 
    COUNT(DISTINCT S.Siting) AS [# Of Types],

    --STRING_AGG(S.Siting,',') AS Animals
    (
        SELECT STRING_AGG(U.SITING, ',')
        FROM 
        (
            SELECT DISTINCT T.Siting
            FROM Sitings AS T
            WHERE 
                T.State = S.State AND
                T.City = S.City
        ) AS U
    ) AS ANIMAL

FROM 
    Sitings AS S
GROUP BY 
    S.State, 
    S.City
ORDER BY
    S.State, 
    S.City

Answer 4

这是另一种方法 (sql fiddle)：

  WITH Sitings 
  AS
  (
    SELECT * FROM (VALUES 
      (1, 'Florida', 'Orlando', 'bird'),
      (2, 'Florida', 'Orlando', 'dog'),
      (3, 'Arizona', 'Phoenix', 'bird'),
      (4, 'Arizona', 'Phoenix', 'dog'),
      (5, 'Arizona', 'Phoenix', 'bird'),
      (6, 'Arizona', 'Phoenix', 'bird'),
      (7, 'Arizona', 'Phoenix', 'bird'),
      (8, 'Arizona', 'Flagstaff', 'dog')
    ) F (ID, State, City, Siting)
  ) 

select State,City,count(*) as [# Of Sitings],(select string_agg(value,', ') from (select distinct value from string_split(string_agg(Siting, ','),',')) t) AS Animals
FROM Sitings 
GROUP BY State, City

您可以轻松地将拆分和合并部分转换为可重用的标量值函数。

注意

这不是最佳解决方案，如果您先分组然后进行聚合（如上面的答案），效果会更好。此外，它不会得到 # of Types，而是得到 # of Sitings。但是，作为快速内联函数，它变得很方便。

在 STRING_AGG 中生成 DISTINCT 值

Produce DISTINCT values in STRING_AGG

sql

sql-server

distinct

sql-server-2017

string-aggregation